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a) Let $T$ be a non zero element of $∧^k (V^*)$ where $\dim⁡ V=k$. Prove that 2 ordered bases $\{v_1,…,v_k \}$ and $\{ v_1' ,…,v_k' \}$ for $V$ are equivalent oriented if and only if $T(v_1,…,v_k )$ and $T(v_1',…,v_k' )$ have the same sign.

b) Suppose that $V$ is oriented. Show that the one dimensional vector space $∧^k (V^*)$ acquires a natural orientation, by defining the sign of a nonzero element $T∈∧^k (V^* )$ to be the sign of $T(v_1,…,v_k )$ for any positive oriented ordered basis ${v_1,…,v_k }$ for $V$

c) Conversely show that an orientation of $∧^k (V^* )$ naturally defines an orientation on $V$ by reserving the above.

Here is what I got so far

a)

=>

If 2 ordered bases $\{v_1,…,v_k \}$ and $\{ v_1' ,…,v_k' \}$ for $V$ are equivalent oriented then $T(v_1,…,v_k )$ and $T(v_1',…,b_k' )$ have the same sign. This is trivial.

<=

Assume that $T(v_1,…,v_k )$ and $T(v_1',…,b_k' )$ have the same sign. let $A=\{v_1,…,v_k \}$ and $B=\{ v_1' ,…,v_k' \}$ By the determinant theorem, $A*T=(det A)T$ for every $T \in \wedge^k(V)$ where $k=dim V$. Since 2 tensors have the same sign, the $det$ of the change of basis matrix are the same, so $A$ and $B$ must be equivalent.

b)

Suppose that $V$ is oriented and there exist a nonzero element $T∈∧^k (V^* )$ to be the sign of $T(v_1,…,v_k )$ for any positive oriented ordered basis ${v_1,…,v_k }$ for $V$, then such $T$ is a basis for $\wedge^k(V^*)$. Call it positive oriented if $T(positive basis)>0$. This goes both way so we can also get c) from this.

this is how I understand the problem. I would be appreciated if anyone can help me improve my argument.

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  • $\begingroup$ how do you define the exterior algebra? $\endgroup$ – hjhjhj57 Dec 10 '14 at 11:20
  • $\begingroup$ exterior algebra is the algebra of wedge product, right? $\endgroup$ – Diane Vanderwaif Dec 10 '14 at 13:15
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Let $\dim(V)=n\geq1$, and assume that the base field is ${\mathbb R}$ or ${\mathbb C}$.

The things you have to prove are almost trivial once you have fully grasped that the space $\wedge^n(V^*)$ of skew $n$-linear forms on $V$ is one-dimensional. When $T\ne0$ is such a form and $A:\>V\to V$ a linear map then the function $T_A$ defined by $$T_A(v_1,\ldots,v_n):=T(Av_1,\ldots, Av_n)\tag{1}$$ is easily seen to be such a form as well. It follows that there is a constant $\lambda_A$ such that $T_A=\lambda_A T$. One immediately verifies that the constant $\lambda_A$ is independent of the chosen $T\ne0$; therefore it is allowed to call it the determinant of $A$, and to denote it by $\det(A)$. It follows that we can replace $(1)$ by $$T(Av_1,\ldots, Av_n)=\det(A)\>T(v_1,\ldots, v_n)\qquad \forall T, \quad \forall A,\quad\forall (v_1,\ldots, v_n)\ .\tag{2}$$ From $(2)$ one concludes that $$\det(AB)=\det(A)\>\det(B)\tag{3}$$ for arbitrary linear transformations $A$, $B$.

I understand that you call two bases $(v_i)_{1\leq i\leq n}$ and $(v'_i)_{1\leq i\leq n}$ equivalent if the map $A$ defined by $Av_i=v'_i$ $\>(1\leq i\leq n)$ has positive determinant. The identity $(3)$ guarantees that this is indeed an equivalence relation. From $(2)$ it then follows that $$T(v'_1,\ldots, v'_n)=T(Av_1,\ldots, Av_n)=\det(A)\>T(v_1,\ldots, v_n)\ ,\tag{4}$$ which immediately implies your claim a).

Claim b) is now equally obvious: If $T(v_1,\ldots,v_n)$ is positive (resp., negative) for one positively oriented basis then it is positive (resp., negative) for any such basis – just because the factor $\det(A)$ in $(4)$ is $>0$.

In order to prove claim c) choose an orientation in $\wedge^n(V^*)$ by declaring a certain $T_0\in \wedge^n(V^*)$ of your choice as positively oriented. It is then easy to see that $${\rm orientation}(v_1,\ldots, v_n):={\rm sgn}\bigl(T_0(v_1,\ldots, v_n)\bigr)$$ defines an orientation on the set of bases of $V$.

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